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miss Akunina [59]

3 years ago

10

students in sixth grade class were asked which team was their favorite. the results are shown in the table. find the ratio of st

udents who preferred the buccaneers to the total number of responses. tigers 35 pathers 45 buccaneers 74 and aviators 52

1 answer:

Serjik [45]3 years ago

8 0

I don't know because I'm not in six grade

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in January, there were 71 cars were made in February another 28 , red cars were made. How many dragon were born altogether show

dolphi86 [110]

Answer:

36

Step-by-step explanation:

7 0

2 years ago

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

2 years ago

IM DOING A TEST!<br><br> Idk how to do this help<br> Please try

goldenfox [79]

Answer:

1 is 102

2 is 78

~~~~~~~~~~~~~~~~~~~~

Always double check :)

7 0

3 years ago

Read 2 more answers

Will you kill baby Hitler

Bad White [126]

Answer:

omg

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Paul and Jim work at a t-shirt factory. They pack t-shirts in boxes and send them

Gwar [14]

Answer:

Jim's box hold more t-shirts by 3 cubic feet

Step-by-step explanation:

Jim's box length = 2 feet

Jim's box breadth = 4 feet

Jim's box height = 6 feet

Volume of box = $Length \times Breadth \times Height = 2 \times 4 \times 6 = 48 feet^3$

Paul's box length = 3 feet

Paul's box breadth =5 feet

Paul's box height = 3 feet

Volume of box = $Length \times Breadth \times Height =3 \times 5 \times 3 = 45 feet^3$

So,Jim's box hold more t shirts by (48-45 = 3 cubic feet )

So, Option C is true

Jim's box hold more t-shirts by 3 cubic feet

3 0

2 years ago